Write the following quadratic function in vertex form. State the x and y-intercepts and the range. 𝑦 = −4𝑥^2 + 20𝑥 − 17
wrong again, on the x-intercepts!
y = −4𝑥^2 + 20𝑥 − 17 = -(2x-5)^2 + 8
so it opens downward, with vertex at (5/2,8)
the x-intercepts, using the quadratic formula,are
x = 5/2 ± √2
You are correct, I apologize for the mistake.
To find the x-intercepts, we can set the function equal to zero and solve for x:
0 = -4𝑥^2 + 20𝑥 - 17
Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a),
we have a = -4, b = 20, and c = -17.
Plugging in these values, we get:
x = (-20 ± √(20^2 - 4(-4)(-17)))/(2(-4))
x = (-20 ± √(400 - 272))/(-8)
x = (-20 ± √128)/(-8)
x = (-20 ± √(64*2))/(-8)
x = (-20 ± 8√2)/(-8)
x = 5/2 ± √2
Therefore, the x-intercepts are x = 5/2 + √2 and x = 5/2 - √2.
Thank you for catching that mistake!
To write the given quadratic function in vertex form, we need to complete the square. The vertex form of a quadratic function is given by 𝑦 = 𝑎(𝑥 − ℎ)^2 + 𝑘, where (ℎ, 𝑘) represents the vertex of the parabola.
Let's begin by rewriting the given quadratic function 𝑦 = −4𝑥^2 + 20𝑥 − 17:
𝑦 = −4𝑥^2 + 20𝑥 − 17
To complete the square, we need to focus on the quadratic term (−4𝑥^2 + 20𝑥). We can factor out a common factor of −4:
𝑦 = −4(𝑥^2 − 5𝑥) − 17
Now we need to find the value to add inside the parentheses to make a perfect square trinomial. To do this, we take half of the coefficient of the linear term (5/2), square it ((5/2)^2 = 25/4), and add it inside the parentheses. However, we also need to consider that we're multiplying the entire term by −4, so we need to subtract (−4)(25/4) = −25 inside the parentheses:
𝑦 = −4(𝑥^2 − 5𝑥 + 25/4 - 25/4) - 17
Simplifying inside the parentheses:
𝑦 = −4(𝑥^2 − 5𝑥 + 25/4) + 25 - 17
𝑦 = −4(𝑥 - 5/2)^2 + 8
Therefore, the quadratic function 𝑦 = −4𝑥^2 + 20𝑥 − 17 can be written in vertex form as 𝑦 = −4(𝑥 - 5/2)^2 + 8.
Now, let's determine the x-intercepts, y-intercept, and the range:
To find the x-intercepts, we set y = 0 and solve for x:
0 = −4(𝑥 - 5/2)^2 + 8
4(𝑥 - 5/2)^2 = 8
(𝑥 - 5/2)^2 = 2
Taking the square root of both sides (considering both positive and negative roots):
𝑥 - 5/2 = ±√2
𝑥 = 5/2 ± √2
So the x-intercepts are 𝑥 = 5/2 + √2 and 𝑥 = 5/2 - √2.
To find the y-intercept, we set x = 0 and solve for y:
𝑦 = −4(0)^2 + 20(0) − 17
𝑦 = −17
Therefore, the y-intercept is 𝑦 = −17.
For the range, notice that the coefficient of the 𝑥^2 term (−4) is negative, indicating that the parabola opens downward. Since the vertex is at the highest point of the parabola, its y-coordinate (8) represents the maximum value of y. So the range of the quadratic function is 𝑦 ≤ 8.
In summary:
- The quadratic function in vertex form is 𝑦 = −4(𝑥 - 5/2)^2 + 8.
- The x-intercepts are 𝑥 = 5/2 + √2 and 𝑥 = 5/2 - √2.
- The y-intercept is 𝑦 = −17.
- The range of the quadratic function is 𝑦 ≤ 8.
To convert the quadratic function 𝑦 = −4𝑥^2 + 20𝑥 − 17 into vertex form, we need to complete the square.
First, let's factor out the common factor of -4 from the first two terms:
𝑦 = -4(𝑥^2 - 5𝑥) - 17
Next, we will complete the square by taking half of the coefficient of 𝑥 (-5), squaring it and adding it inside the parentheses:
𝑦 = -4(𝑥^2 - 5𝑥 + 6.25) - 17 - 4(6.25)
Now, let's simplify:
𝑦 = -4(𝑥 - 2.5)^2 - 17 - 25
Combining like terms, we have:
𝑦 = -4(𝑥 - 2.5)^2 - 42
Therefore, the quadratic function 𝑦 = −4𝑥^2 + 20𝑥 − 17 can be written in vertex form as 𝑦 = -4(𝑥 - 2.5)^2 - 42.
To find the x-intercepts, we set 𝑦 equal to zero and solve the resulting quadratic equation:
0 = -4(𝑥 - 2.5)^2 - 42
4(𝑥 - 2.5)^2 = -42
(𝑥 - 2.5)^2 = -10.5
Since the square of any real number is always greater than or equal to zero, there are no x-intercepts for this quadratic function.
The y-intercept occurs when 𝑥 is equal to zero. Plugging 𝑥 = 0 into the quadratic function, we get:
𝑦 = -4(0)^2 + 20(0) - 17
𝑦 = -17
Therefore, the y-intercept is -17.
To determine the range of the function, let's analyze the coefficient of the squared term (-4). Since the coefficient is negative, the parabola opens downward. This means that the y-values will be negative or zero for all possible x-values. Hence, the range of the function is (-∞, 0].